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goldenfox [79]
2 years ago
14

A sound wave, generated at a frequency of 440 hertz has a wavelength of 2.3 meters as it travels through a solid material. The a

pproximate speed of the wave is ____ m/sec.
Physics
1 answer:
Jlenok [28]2 years ago
8 0

The approximate speed of the sound wave traveling through the solid material is 1012m/s.

<h3>Wavelength, Frequency and Speed</h3>

Wavelength is simply the distance over which the shapes of waves are repeated. It is the spatial period of a periodic wave.

From the wavelength, frequency and speed relation,

λ = v ÷ f

Where λ is wavelength, v is velocity/speed and f is frequency.

Given the data in the question;

  • Frequency of sound wave f = 440Hz = 440s⁻¹
  • Wavelength of the wave λ = 2.3m
  • Speed of the wave v = ?

To determine the approximate speed of the wave, we substitute our given values into the expression above.

λ = v ÷ f

2.3m = v ÷ 440s⁻¹

v = 2.3m × 440s⁻¹

v = 1012ms⁻¹

v = 1012m/s

Therefore, the approximate speed of the sound wave traveling through the solid material is 1012m/s.

Learn more about Speed, Frequency and Wavelength here: brainly.com/question/27120701

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A conducting spherical shell with inner radius a and outer radius b has a positive point charge +Q at the center in the empty re
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Answer:

a) q_inner = -Q

, q_outer = -2Q

b)    E₁ = k Q / r²        r<a

       E₂ = 0               a<r<b

       E₃ = - k 2Q/r²     r>b

d)   the charge continues inside the spherical shell, the results do not change

Explanation:

a) The point load in the center induces a load on the inner surface of the shell with constant opposite sign  

q_inner = -Q  

the outer shell of the shell the load is  

q_outer = -3 Q + Q  

q_outer = -2Q

b) To find the electric field again, use Gauss's law,  

We define as a Gaussian surface a sphere  

Ф = E. dA = q_{int}/ε₀

in this case the electric field lines and the radii of the sphere are parallel, so the sclar product is reduced to the algegraic product  

E A = q_{int}/ε₀

the area of ​​a sphere is  

A = 4 π r²  

E = 1 / 4πε₀  Q/ r²  

k = 1 / 4πε₀  

let's apply this expression to the different radii

i) r <a  

in this case the load inside is the point load  

q_{int}= + Q  

E₁ = k Q / r²

ii) the field inside the shell  

a <r <b  

As the sphere is conductive, so that it is in electrostatic equilibrium, there can be no field within it.  

E₂ = 0

iii)  r>b

   q_{int}  = Q- 3Q = -2Q

    E₃ = k (-2Q/r²)

     E₃ = - k 2Q/r²

c) see  attached

d) as the charge continues inside the spherical shell, the results do not change, since the lcharge inside remains the same and it does not matter its precise location, but remains within the Gaussian surface

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3 years ago
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